I am a beginner of digital signal processing. In my research project, I need to calculate this equation which is below.

$$ e(t) = \Re \Big\{z(t) \, e^{j \omega_0 t} \Big\} $$

I know the $e(t)$ and I want to know how to calculate $z(t)$ and $\omega_0$. And I also want to know what's the meaning of $\omega_0$, if it's the same as the center frequency of $e(t)$.

I am sorry about that this question is very basic. Thanks!


1 Answer 1


If $e(t)$ is band-limited and if the frequency $\omega_0$ is chosen such that the complex signal

$$y(t)=z(t)e^{j\omega_0 t}\tag{1}$$

has no negative frequencies, then $y(t)$ is called the analytic signal, and it can be determined from $e(t)$ using the Hilbert transform:


This analytic signal $y(t)$ is unique. However, the complex envelope $z(t)$ is not, unless we define the frequency $\omega_0$. Even though $\omega_0$ can be called center frequency, there is no unique definition of such a center frequency. That's why $\omega_0$ is often called reference frequency, and the complex envelope $z(t)$ depends on the choice of that reference frequency.

If $\omega_0$ is not given, you cannot uniquely solve for $z(t)$. It's up to you to define a convenient value for $\omega_0$, from which the complex envelope is easily obtained:

$$z(t)=y(t)e^{-j\omega_0 t}\tag{3}$$

Usually you would choose $\omega_0$ such that $z(t)$ is a low pass signal.

Note that since

$$e(t)=\Re\{z(t)e^{j\omega_0 t}\}=\frac12\left\{z(t)e^{j\omega_0 t}+z^*(t)e^{-j\omega_0 t}\right\}\tag{4}$$

(where $*$ denotes complex conjugation), the complex envelope can also be determined form $e(t)$ by down conversion and low pass filtering:

$$2e(t)e^{-j\omega_0 t}=z(t)+z^*(t)e^{-2j\omega_0 t}\tag{5}$$

Assuming that $z(t)$ is a low pass signal and that $\omega_0$ is chosen such that it is greater than the bandwidth of $z(t)$, low pass filtering $(5)$ will yield $z(t)$.


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